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I was following through The Secret Life Of Equations by Rich Cochrane, and in the chapter on Newton's Second Law, there is a example: given an object of mass $1\, kg$, thrown up at $20\, m/s$ by someone $2$ metres tall, how high does it get?

It starts by creating function $h(t)$ that calculates the height of the object at time t. It then puts: $-9.8 = \frac{d^2h}{dt^2}$, showing the acceleration of the object. Then it integrates: $-9.8t + C = \frac{dh}{dt}$, but in this case know C, it equals the start acceleration of the object, so $-9.8t + 20 = \frac{dh}{dt}$. At this point it calculates the time when $-9.8t + 20 = 0$, which is the time the object reaches its maximum height. The book says it's around 2 seconds (the actually value is 2.04). Then it integrates once more: $-9.8t^2+20t+C = h$, but again we know C! It's the starting height of the object, so $-9.8t^2+20t+2=h$. When we put in the previously calculated value of t at the maximum height, 2, in we get : $22.4$ metres. Then, coming around to my actually question, the book says that "with a little more algebra, we can calculate that I'll catch the ball around 4 seconds after it left my hand". But surely this is not a complex calculation, because we are not factoring in air resistance, so $\mathrm{falling\ time} = \mathrm{rising\ time}$ so total time in air equals twice the rising time.

So my question is: why would you choose to recalculate time in air from maximum height rather than just double rising time? Consider this is on a flat earth with constant gravity and no air resistance.

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  • $\begingroup$ You can double the time or recalculate, it will give the same results $\endgroup$
    – Ofek Gillon
    Commented Feb 27, 2017 at 20:58
  • $\begingroup$ Time reversal symmetry means that going up = going down if there are no dissipative forces (e.g. air resistance) involved. $\endgroup$
    – Jon Custer
    Commented Feb 27, 2017 at 21:45
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    $\begingroup$ Your only reason for asking this question seems to be to point out the deficiency in the book and to ask people to agree with you that the author made the missed an obvious shortcut. $\endgroup$
    – sammy gerbil
    Commented Feb 28, 2017 at 2:15
  • $\begingroup$ @sammygerbil No, I actually wanted to find out if the way the book did it was more accurate, or perhaps easier. $\endgroup$
    – user146934
    Commented Feb 28, 2017 at 8:05
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    $\begingroup$ That is not the impression your question gives, eg "Surely this is not a complex calculation..." That is the question or a critic. Your comments about air resistance show that you understand the physics of the situation very well. Why might you think the book's method is more accurate? It is just the same calculation done a 2nd time, as you are aware. $\endgroup$
    – sammy gerbil
    Commented Feb 28, 2017 at 8:16

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When there is no drag, you do not need to recalculate time, you just double it. A simple explanation why this is true is based on energy conservation.

Since there is no dissipative forces the mechanical energy, which is the sum of the kinetic and potential energies, is constant. The potential energy depends only the height so at any height the body has the same kinetic energy no matters it is going up or down. Here, same kinetic energy means same speed and therefore the time of raising is the same of falling.

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